<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\un}{\underline}
\usepackage{epsfig}
\begin{document}
\parindent=0pt

\textbf{Question}

A solid sphere $r=a$ is held fixed in an otherwise uniform free stream
of speed $U$ which flows parallel to the $x$-axis. The REynolds number
of the stream is small. The fluid has constant density $\rho$ and
constant kinematic viscosity $\nu$ and its velocity it denoted by
$\un{q} = u \un{e}_r + v \un{e}_{\theta}$ where $\un{e}_r$ and
$\un{e}_{\theta}$ are unit vectors in the $r$ and $\theta$ directions
respectively and $(t, \theta, \phi)$ are spherical polar
coordinates. The flow is axisymmetric (i.e. independent of $\phi$).

YOU MAY ASSUME that the Stoke stream function $\psi (r, \theta)$ for
the flow is defined by
\begin{eqnarray*}
u &amp; = &amp; \frac{1}{r^2\sin\theta} \frac{\partial \psi}{\partial
\theta}\\
v &amp; = &amp; -\frac{1}{r\sin \theta} \frac{\partial \psi}{\partial r}
\end{eqnarray*}
and that the slow flow equations in spherical polar coordinates are
$$\left ( \frac{\partial^2}{\partial^2 r} + \frac{1}{r^2}
\frac{\partial^2}{\partial \theta^2} - \frac{\cot\theta}{r^2}
\frac{\partial}{\partial \theta} \right )^2 \psi = 0.$$

\begin{description}
%Question 5i
\item{(i)}
Determine the boundary conditions for the flow.

%Question 5ii
\item{(ii)}
By assuming a stream function of the form
$$\psi = U g(r)\sin^2 \theta$$
obtain $u$ and $v$

%Question 5iii
\item{(iii)}
Sketch the flow streamlines.

%Question 5iv
\item{(iv)}
By considering the orders of magnitude of $\un{q}$ for large calues of
a suitably non-dimensionalised $r$, show that slow flow theory (and
hence the analysis carried out above) is invalid if $r \ge O(1/Re)$.
\end{description}

\newpage
\textbf{Answer}

\begin{description}

%Question 5i
\item{(i)} at $r=a$ need $\psi=\psi_r=0$ (no-slip).

As $r\to\infty$ we have $\underline{q}=U\hat{e}_x$.

Now $\hat{e}_r=\cos\theta\hat{e}_x+\sin\theta\hat{e}_y$, \ \
$\hat{e}_{\theta}=-\sin\theta\hat{e}_x+\cos\theta\hat{e}_y$

$\Rightarrow \hat{e}_r-\sin\theta\hat{e}_{\theta}$ $$\displaystyle
\frac{1}{r^2\sin\theta}\psi_{\theta}=U\cos\theta, \ \ \
\frac{-1}{r\sin\theta}\psi_r=-U\sin\theta$$

$\Rightarrow \psi_r\sim\frac{1}{2}Ur^2\sin^2\theta$ as
$r\to\infty$

Now try $\psi=Ug(r)\sin^2\theta$

%Question 5ii
\item{(ii)}
$\displaystyle \left ( \frac{\partial^2}{\partial
r^2}+\frac{1}{v^2}\frac{\partial^2}{\partial
\theta^2}-\frac{\cot\theta}{r^2}\frac{\partial}{\partial\theta}
\right )Ug(r)\sin^2\theta$

$\displaystyle =U\left [ g''sin^2\theta
+\frac{g}{r^2}(2\cos^2\theta-2\sin^2\theta)-\frac{\cos\theta}{r^2\sin\theta}g2\sin\theta\cos\theta
\right ] $

$\displaystyle =U\left [
g''\sin^2\theta-\frac{2g}{r^2}\sin^2\theta \right ] =
U\sin^2\theta \left [ g''-\frac{2g}{r^2} \right ] $

Applying the operator again$\Rightarrow$

\begin{eqnarray*}
&amp; &amp; U\sin^2\theta \left ( g''-\frac{2g}{r^2} \right
)''+\frac{U}{r^2} \left ( g''-\frac{2g}{r^2} \right )
(2\cos^2\theta-2\sin^2\theta)\\
&amp; &amp; -\frac{\cot\theta}{r^2} \left (
g''-\frac{2g}{r^2} \right ) 2U\sin\theta\cos\theta\\
&amp; = &amp;U\sin^2\theta \left ( g'''- \left
(\frac{v^22g'-2g2r}{r^4} \right ) \right ) '\\
&amp; &amp;  -2 \left (
\frac{g''}{r^2}-\frac{2g}{r^4} \right ) U \sin^2\theta = 0\\
&amp; = &amp; U\sin^2\theta \left ( g'''' - \left (
\frac{2g'}{r^2} \right )' + \left ( \frac{4g}{r^3} \right ) '
-\frac{2g''}{r^2} + \frac{4g}{r^4} \right ) = 0\\
&amp; = &amp; U\sin^2\theta \left ( g'''' - \left (
\frac{r^22g''-2r2g'}{r^4} \right ) \right.\\
&amp; &amp; + 4 \left. \left (
\frac{r^3g'-g3r^2}{r^6} \right ) - \frac{2g''}{r^2} +
\frac{4g}{r^4} \right )\\
&amp; = &amp; 0\\
\Rightarrow &amp; &amp; g''''-\frac{4g''}{r^2}+\frac{8g'}{r^3}
-\frac{8g}{r^4} = 0, \ \ \ \rm{put\ } g=r^n
\end{eqnarray*}


$n(n-l)(n-2)(n-3)-4n(n-1)+8n-8=0$

$(n-1)[n(n-2)(n-3)-4n+8]=0$

$(n-1)(n-2)[n^2-3n-4]=0$, \ \ \ $n=1,2,-1,4$

$\displaystyle \Rightarrow \psi=U\sin^2\theta \left [
Ar^4+Br^2+Cr+\frac{D}{r} \right ] $

Conditions at $\infty$, \ $\Rightarrow A=0,\ \ B=\frac{1}{2}$

Conditions at $r=a$ $\Rightarrow$ $\left. \begin{array}{r}
\displaystyle \frac{a^2}{2}+Ca+\frac{D}{a}=0\\ \displaystyle
a+c-\frac{D}{a^2}=0
\end{array} \right \}$ $\left.
\begin{array}{l}\displaystyle C=\frac{-3a}{4}\\ \displaystyle D=\frac{a^3}{4} \end{array}
\right.$

$\Rightarrow$ $\displaystyle \psi=U\sin^2\theta \left [
\frac{r^2}{2}-\frac{3ar}{4}+\frac{a^3}{4r} \right ]$

\begin{eqnarray*} \Rightarrow u &amp; = &amp; U\cos\theta \left [
\displaystyle 1-\frac{3a}{2r}+\frac{a^3}{2r^3} \right ] \\ v &amp; = &amp;
U\sin\theta \left [ -1+\frac{3a}{4r}+\frac{a^3}{4r^3} \right ]
\end{eqnarray*}

%Question 5iii
\item{(iii)}

\begin{center}
\epsfig{file=311-flowpastsphere.eps, width=90mm}
\end{center}

(Any reasonable symmetric effort will be accepted)

%Question5iv
\item{(iv)} We have ignored$Re(\underline{q}.\nabla)\underline{q}$
in comparision to $\nabla^2\underline{q}$. Now for large $r$,
$\underline{q}\sim 1$, $\nabla\sim\frac{1}{r}$ \ \
(non-dimensional).

$\Rightarrow Re\left ( \frac{1}{r} \right ) \ll \frac{1}{r^2}$ \ \
\ if slow flow is to hold.

$\Rightarrow Re\ll \frac{1}{r}$. \ \ But for fixed $Re$, (however
small) we can always choose r large enough to violate this.

$\Rightarrow$ not valid for $r \ge O\left (\frac{1}{Re} \right )$

\end{description}

\end{document}
</pre></body></html>